Derivative of \( \displaystyle \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} \)
Problem 2.65 · hard Beautiful
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)}{10 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(5 x + 1 \right)}}{10 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]sumApply the sum rule to the inner expression.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan^{2}{\left(5 x + 1 \right)}}{10 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]constantThe derivative of a constant is zero.✓ Proved
- \[ = \frac{\tan{\left(5 x + 1 \right)} \frac{d}{d x} \tan{\left(5 x + 1 \right)}}{5 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]powerApply the power rule to the squared term.✓ Proved
- \[ = \frac{\tan{\left(5 x + 1 \right)} \sec^{2}{\left(5 x + 1 \right)} \frac{d}{d x} \left(5 x + 1\right)}{5 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]trigApply the chain rule to the tangent function.≈ Checked numerically
- \[ = \frac{\tan{\left(5 x + 1 \right)} \sec^{2}{\left(5 x + 1 \right)}}{\tan^{2}{\left(5 x + 1 \right)} + 1} \]chain algebraApply the chain rule to the linear term. Simplify the constants.✓ Proved
- \[ = \tan{\left(5 x + 1 \right)} \]rewrite algebra simplifyUse the identity sec(u)**2 = 1 + tan(u)**2. Distribute the division. Cancel the common term in the numerator and denominator.≈ Checked numerically
Answer \( \tan{\left(5 x + 1 \right)} \)
Mind the domain. The answer is also defined at points where f(x) is not. Substituting there gives a number that is not a slope of f.
Lines: 11 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(5*x + 1)**2 - sec(5*x + 1)**2 + 1)*tan(5*x + 1)/(tan(5*x + 1)**2 + 1); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 sec has poles at odd multiples of pi/2 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 10 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-tan(5*x + 1)**2 + sec(5*x + 1)**2 - 1)*tan(5*x + 1)/(tan(5*x + 1)**2 + 1); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: passdeepseek-r1:70b: passqwen3.6:27b-mlx: pass
Every verdict on record (12)
qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: fail (error) 2026-09-20 — Step 9 applies two simplifications at once: it cancels the 1/10 with the 10 and simplifies the fraction, violating the rule that each step must change only one thing. The correct result after step 8 should be (tan(5*x+1)*sec(5*x+1)**2)/(tan(5*x+1)**2+1).qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed, and the final simplification is correct.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: pass 2026-09-19deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19gpt-oss:20b: pass 2026-09-17deepseek-r1:70b: pass 2026-09-17
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-09-26 with SymPy 1.14.0.